# Time reversal symmetry for Zeeman fields

For a magnetic field along $\hat{x}$ the hamiltonian is given by: $H_{z}=V_x(c^{\dagger}_{\uparrow}c_{\downarrow}+c^{\dagger}_{\downarrow}c_{\uparrow})$.

If we follow the usual prescription of changing the spins: $\uparrow\quad \rightarrow \quad\downarrow$ and vice-versa, how do we see that the above hamiltonian breaks TR symmetry?

## 1 Answer

Time reversal sends $c_\uparrow\rightarrow c_\downarrow$ and $c_\downarrow\rightarrow -c_\uparrow$. With this transformation, it should be clear the term is time odd.

See, e.g., here for more information. The key thing to realize is that while time reversal must send $k\rightarrow -k$ and $+z\rightarrow -z$, that still leaves the possibility of gaining an arbitrary phase under time reversal. To determine this phase, you demand that time reversal also sends $+x$ spins to $-x$ spins, and $+y$ spins to $-y$ spins. With these constraints, you can find the phase factors above.

• Is there any reference for this? I thought the procedure only required flipping the spins. (and k -> -k in momentum space) – Arnab Barman Ray May 28 '18 at 16:30